Solve Trig Equation 485E9D

1. **State the problem:** Solve the equation $$\sin x + \cos x + \sin x \cos x = 1$$. 2. **Use the identity:** Recall that $$(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2 \sin x \cos x = 1 + 2 \sin x \cos x$$. 3. **Express $\sin x \cos x$ in terms of $\sin x + \cos x$:** $$\sin x \cos x = \frac{(\sin x + \cos x)^2 - 1}{2}$$. 4. **Substitute into the original equation:** $$\sin x + \cos x + \frac{(\sin x + \cos x)^2 - 1}{2} = 1$$. 5. **Multiply both sides by 2 to clear the denominator:** $$2(\sin x + \cos x) + (\sin x + \cos x)^2 - 1 = 2$$. 6. **Rearrange:** $$ (\sin x + \cos x)^2 + 2(\sin x + \cos x) - 3 = 0$$. 7. **Let $u = \sin x + \cos x$, with $-1 \leq u \leq 1$:** $$u^2 + 2u - 3 = 0$$. 8. **Factor the quadratic:** $$(u - 1)(u + 3) = 0$$. 9. **Solve for $u$:** $$u = 1 \quad \text{or} \quad u = -3$$. 10. **Check domain:** Since $u = \sin x + \cos x$ and $-1 \leq u \leq 1$, only $u = 1$ is valid. 11. **Solve $\sin x + \cos x = 1$:** Rewrite as: $$\sqrt{2} \sin \left(x + \frac{\pi}{4}\right) = 1$$ $$\sin \left(x + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$$. 12. **Find principal value:** $$\text{pv} = \sin^{-1} \left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$$. 13. **General solution for sine equation:** $$x + \frac{\pi}{4} = \pi n + (-1)^n \frac{\pi}{4}, \quad n \in \mathbb{Z}$$ 14. **Isolate $x$:** $$x = -\frac{\pi}{4} + (-1)^n \frac{\pi}{4} + \pi n, \quad n \in \mathbb{Z}$$. **Final answer:** $$x = -\frac{\pi}{4} + (-1)^n \frac{\pi}{4} + \pi n, \quad n \in \mathbb{Z}$$